Edit: I suddenly found that this proof is called Abel Identity. And the proof of it done by me is exactly the same as what he did.
At first, I‘d like to say thank you to MIT open courses which give me the privilege to enjoy the most outstanding education resources.
Okay, come to the point. When I was learning the second-order homogeneous differential equation, the professor quoted a theory in one step to prove that ${c_{1}y_{1}+c_{2}y_{2}}$ are all the solutions.
THM: if $y_{1},y_{2}$ are solu‘s to ODE, then either $W(y_{1},y_{2}) = 0$ (i.e. for all x) or$W(y_{1},y_{2})$ is nonzero (i.e. for all x).
note: W means Wronskian
Well, frankly speaking, I got the inspiration from that introduction in wiki too.
Proof:
Let‘s say that we have a second-order homogeneous differential equation.
$y‘‘ + py‘ + qy = 0$ in which $p$ and $q$ are some functions of x.
And $y_{1},y_{2}$ are two solutions to this equation, which means that:
$y_{1}‘‘ + py_{1}‘ + qy_{1} = 0$
$y_{2}‘‘ + py_{2}‘ + qy_{2} = 0$
Last, we get our $W$, $W(y_{1},y_{2}) = y_{1}y_{2}‘ - y_{2}y_{1}‘$.
Step 1. Differential $W$, and we get $W‘ = y_{1}y_{2}‘‘ - y_{2}y_{1}‘‘$.
Step 2. From above, we get $y_{1}‘‘ = -py_{1}‘ -qy_{1} $ and $y_{2}$ is the same case.
Step 3. Plug these in, we get $W‘ = (-p)(y_{1}y_{2}‘ - y_{2}y_{1}‘) = (-p)W$, and that is $W‘ + pW = 0$. This is the first solvable linear equation.
Step 4. So we get $W = ce^{-\int p(x)\,dx}$, and $c$ is an arbitrary constant. As a result, if $c$ equals to zero, then $W$ is always zero. If $c$ doesn‘t equal to zero, then $W$ is a exponential function, which means that it always $>0$.
Done.
原文：https://www.cnblogs.com/raymondjiang/p/11622882.html